# Why is angular momentum conserved in a satellite?

Mar 8, 2018

The principle of conservation of momentum, both linear (also known as translational) and angular, is a universal principle of Physics.

#### Explanation:

Angular momentum, $L$, is defined, in mathematical terms as

$L = I \cdot \omega$

Take the time derivative of that

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

We give the term "torque" to $I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}}$ and we give torque the variable name $\tau$.
We give $\frac{\mathrm{do} m e g a}{\mathrm{dt}}$ the variable name $\alpha$.

Expanding that last equation with the variable names $\alpha \mathmr{and} \tau$

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}} = I \cdot \alpha = \tau$ ... Equation1

The equation $\tau = I \cdot \alpha$ is the angular equivalent of $F = m \cdot a$ which Newton made famous in Newton's 2nd Law.

$\tau , I , \mathmr{and} \alpha$ are the angular equivalents of $F , m , \mathmr{and} a$ respectively.

If there are no external torques acting on a body, $\tau = 0$, and that there is no angular acceleration follows. That is shown in the repeat of Equation 1 below.

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}} = I \cdot \alpha = 0$

Therefore $\frac{\mathrm{dL}}{\mathrm{dt}} = 0$. And therefore, the angular momentum is constant if $\tau = 0$.

Therefore, we have Conservation of Angular Momentum .

I hope this helps,
Steve